Solve the equations
m2 + n2 = 29
m + n = 7
(2, 3) and ( 3, 5)
(2, 5) and (5, 2)
(5, 2) and ( 5, 3)
(5, 3) and (3, 5)
Correct answer is B
m2 + n2 = 29 .......(1)
m + n = 7 ............(2)
From (2),
m = 7 - n
but m2 + n2 = 29, substituting;
(7-n)2 + n2 = 29
49 - 14n + n2 + n2 = 29
=> 2n2 -14n + 20 = 0
Thus n2 -7n + 10 = 0
Factorizing;
(n-5)(n-2) = 0
n - 5 = 0, => n = 5
n - 2 = 0, => n = 2.
When n = 5,
m + n = 7, => m = 2,
When n = 2,
m + n = 7, => m = 5.
Thus (m,n) = (5,2) and (2,5)
(1,1), (1,2), (1,3)
(1,1), (2,1), (1,3)
(1,1), (3,1), (2,2)
(1,1), (1,2), (2,1)
Correct answer is D
Hint: Sketch the inequality graph for the 3 conditions given and read out your points from the co-ordinates.
\(\frac{q-8}{7}\)
\(\frac{7-q}{4}\)
\(\frac{8-q}{2}\)
\(\frac{7+q}{8}\)
Correct answer is B
\(y = px^{2} + q ... (i)\)
\(y = 2x^{2} - 1 ... (ii)\)
At x = 2,
(i): \(y = p(2^{2}) + q = 4p + q\)
(ii): \(y = 2(2^{2}) - 1 = 7\)
\(\therefore \text{The coordinates of the point of intersection = (2, 7)}\)
(i): \(7 = 4p + q \implies p = \frac{7 - q}{4}\)
Divide: \(a^{3x} - 26a^{2x} + 156a^{x} - 216\) by \(a^{2x} - 24a^{x} + 108\).
ax - 2
ax + 2
ax - 8
ax - 6
Correct answer is A
\(\frac{a^{3x} - 26a^{2x} + 156a^{x} - 216}{a^{2x} - 24a^{x} + 108}\)
Let \(a^{x} = z\)
\(\therefore = \frac{z^{3} - 26z^{2} + 156z - 216}{z^{2} - 24y + 108} ... (i)\)
Dividing (i) above, we get \(z - 2\)
= \(a^{x} - 2\)
Evaluate \(\frac{(0.14^2 \times 0.275)}{7(0.02)}\) to 3 decimal places.
0.039
0.385
0.033
0.038
Correct answer is A
\(\frac{(0.14)^{2} \times 0.275}{7(0.02)} = \frac{(0.14)^{2} \times 0.275}{0.14}\)
= \(0.14 \times 0.275\)
= \(0.0385 \approxeq 0.039\)
Note: Embedded zeroes are counted just the same as non-zero digits and For the number of decimal places stated, count that number of digits to the right of the decimal and underline it.
The next number to its right is called the 'rounder decider'. If the 'rounder decider' is 5 or more, then round the previous digit up by 1